After I have calculated my estimate of the height of the tower, I will be able to begin to calculate the average velocities throughout the stated stages of the ride. I will calculate the average velocity throughout various stages of the ride, these are:
- The journey to the top of the ride
- The journey to the bottom of the ride
- The journey for the whole ride
- The distance of the free fall
- The braking distance
To calculate this I used a stopwatch, which enabled me to time the different stages of the ride. After finding this value I can find the average velocity, but to do this I will use the following formulae:
After analysing my results I found that my results are inequitable as I haven’t taken into consideration of the following:
- Friction
- Air Resistance
- Human reaction to stopping/starting stop watch
Although as the values of these are quite small they are small enough to be ignored.
I assessed the free fall by estimating the displacement from where the carriage is released from the top to where it begins to brake, just before the magnetic brakes. As I said before with any estimation there will be errors. As I calculated the freefall through inspection the estimation isn’t going to be exact making my results inaccurate. After researching on the internet I was later to find that the free fall distance is 29.5meters, only 2.5meters out. Whilst researching I also found that the exact free fall velocity value was 15ms-1, again making my results slightly inaccurate by 3.36ms-1. Due to using a stopwatch and inspection, the errors are most likely to be human. For example - the delay in stopping the stopwatch when required.
Time
I obtained these results with a digital stopwatch. In order to get the best results I worked out an average time for each stage of the ride stated. I collected 5 results for each section and worked out the average to give me a more accurate reading. Human error could also play a part in the inaccuracy of these results.
Below is the results for the time, velocity and displacement for each stage of the ride.
Free fall
Galileo was the man who discovered the theory known as free fall. Galileo discovered this by dropping balls of different mass from the Leaning Tower of Pisa. This led to the conclusion that all objects free fall at the same rate, despite their mass. Anything in free fall is only subjected by the force of gravity which is -9.81ms-2.
I will use the following formula to see whether the detonator’s acceleration is -9.81ms:
All I have to do is insert the known values to find my answer.
- Acceleration (a)=?
- Displacement (S)=27meters
- Time (t)= 2.32 seconds
- Initial velocity (u)= 0ms-1
- Final velocity (v)= n/a
S= ut + 1/2at2
S = 1/2 at2
a= 2s/t2
a = 2x 27
2.322
a= 10.03 ms-2
As you can see my result is near to that of Galileo’s, but Because of things like human error and estimations my result is 0.21ms-2 out. Taking into consideration all the inaccuracies I can safely say that the carriage free falls at the same acceleration as that of the gravitational pull of the earth.
Newton’s Law
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Newton later took Galileo's discoveries concerning mechanics and incorporated these principles into his laws of motion. Newton’s first law applied very much to the ride which states; “every body remains at rest or in uniform motion in a straight lie unless acted upon by a force.” It is this first law of motion, which applies to the ride, firstly when the ride is at the top and is stationary. Whilst at the top there are two forces acting upon the seat of the carriage. The forces act in such a way that they counteract to each other. As the rider sits in their seat at this moment, the seat pushes upward with a force equal in strength and opposite in direction to the force of gravity, which is causing the passengers weight. These two forces are said to balance each other, causing the rider to remain at rest. If the seat is suddenly pulled out from under the rider, then he experiences an unbalanced force. There is no longer an upward seat force to balance the downward pull of gravity, so he accelerates uniformly to the ground, until the carriage decelerates. It is at this point where a force is applied to slow the carriage down bringing it to a controlled stop.
The detonator can give its passengers the sensation of free-fall. It consists of three distinct parts: ascending to the top, momentary suspension, and then the quick downward plunge.
Climb upward
In the first part of the ride, force applied to the car lifts it to the top of the tower. The amount of force depends on the mass of the car and the passengers within the car. Motors create this upward force. There are built-in safety allowances for variables concerning the mass of the riders. F=ma, F=2580x-9.81, F= 25309.8N. This may not be a very accurate result as the mass of the carriage is just an estimate and not a true value. Whilst the carriage is gaining height it is also gaining gravitational energy. The carriage also contains little kinetic energy on its journey.
Transitory suspension
Whilst the car is stationary at the top of the tower the gravitational potential energy is at its greatest. Gravitational potential energy is the energy possessed by the carriage because of its height above the ground. The amount of GPE depends upon its mass and its height. I have worked the GPE with the following formulae:
Gain in gravitational potential energy = mass x gravity x height.
2580(kg) x 9.81(ms-1) x 31.08(m)
= 786628N
The value for the GPE may be inaccurate as I estimated the height and this might not be the exact value. Also the mass of the carriage it an estimate and is not identical for every ride, so it varies for every ride.
With the GPE I am now able to calculate the minimum power used by the motor to lift the carriage. I will use the formula: power=Energy Transferred/Time.
P=786628(N)/17.82(s)
P = 44143 watts
This value may be inaccurate as the GPE I calculated is not an accurate value, also the time may be inaccurate as I used a stopwatch and inaccuracy could have occurred. This is the minimum power used, as the motor is not 100% efficient due to friction, increase/decrease in mass due to passenger mass, air resistance, also loss of power due to heat e.t.c.
Newton’s third law of motion states that; “To every action there is an equal and opposite reaction”. A 6okg man would experience a 0N resultant force acting on his seat whilst at rest, and a 589N reaction force acting on his seat. F=ma, F=60x9.81=589N.
Descending plummet
From the top of the tower the car is swiftly accelerated towards the ground under the influence of the earths gravitational field only, as I previously calculated. It is at this point, which the car gains kinetic energy and looses potential energy. A unique magnetic braking system helps bring the ride to a controlled stop. The faster the carriage moves, the larger resistive forces become.
F=ma
F=2580x9.81=25309.8N
We measure the energy an object has by the work it can do. The work is done when the carriage is moved by the force of gravity. I will now calculate the work done:
work done = force x displacement in direction of force
wd=25310x27
wd= 683370j this is not very accurate as I did not consider air resistance or friction. Also the force involves an estimate of the mass of the carriage, which is not identical in all the rides.
KE: Kinetic energy
The kinetic energy of the carriage is the energy possessed because of its motion. The kinetic energy is greatest when it is at a minimum height. I have worked out the kinetic energy possessed by the carriage whilst in free-fall.
Loss of Gravitational Potential Energy = Gain in kinetic Energy
Mgh = ½mv2
gh = ½ v2
9.81 x 31.08 = ½ v2
304.8 = ½ v2
√609 = v
V = 25ms-1
The main limitation in this calculation is the fact that the height of the ride is an estimate, which may be inaccurate. Also I did not consider the air resistance or friction of the carriage and tower, which would slow the carriage down. In most cases this is small enough to be neglected.
I will now use the constant acceleration formulas to calculate the final velocity just after the free-fall part of the ride, just before the deceleration.
A= -9.8ms-2
S= 27meters
T= 2.32s
U= 0ms-1
V= ?
I will now calculate the rate of deceleration as the carriage comes to a stop. I will use the constant acceleration formulas. Displacement is 4.08 meters as I estimated (page 1), the time for the carriage to stop during deceleration is 2.32 seconds as I also recorded; the initial velocity is 23.3ms-1 as I calculated above.
A=?
S= 4.08m
T= 1.7s
U= 23.3ms-1
V= 0
V2 = u2 + 2as
v2_u 2 = a
2s
a = 02 – 23.32
2x4.08
a=66.5ms-2
=6.8g’s
I looked up onto the Internet and found out that the carriage is changing its speed at a rate of 5g’s. My calculation was 6.8g’s, (6.8 – 5 =) I was just 1.8g’s out. There are many factors, which could have caused this inaccuracy. Firstly the displacement of the deceleration is an estimate, I measured the distance that the carriage travels whilst decelerating the same way I measured the height, by using similar triangles (see page 1). This was an inaccurate measurement as I latter found out on the Internet the displacement to be 5.0 meters, I was 0.92 meters out. This affected my calculation significantly. Secondly the time was such a small measurement that it was really hard for me to get the exact time whilst the carriage is decelerating as I used a manually operated digital stop watch and my reactions were not fast enough. Also I did not take into account air resistance or friction of the carriage, which would slow the carriage down. This is usually a small amount as the mass of the carriage is high but would affect my result slightly.
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The sensation of weightlessness
The sensation of weightlessness is caused by the rider not feeling an outside force upon their bodies. At the free-fall stage, a 60kg rider would feel 589N of force from the seat pushing as an external force upon their body, enabling the rider to feel their normal weight. When the carriage descends the force of the seat is removed, the sensation of weightlessness is achieved. During this brief encounter the rider hasn’t lost any weight, just the absence of the external seat force.
Possible future developments to the ‘Detonator’?
It would be good to see the detonators height increase to maybe double its length. This would enable the rider a long period of weightlessness and make it more thrilling than it already is. Increasing the height would increase the gravitational potential energy making the carriage loose more kinetic energy making the descent quicker and faster.
Investigation Improvements
After concluding my investigation, I’m pleased with the final results. To improve upon this I would have done the following:
- Gathered more time readings to get a better average
- Use computerised equipment to cancel out any human error
- Research how much friction is caused
Bibliography
- www.amusement-park.searchmole.co.uk/ search/amusement-park.html
- A – Z of physic principles
- Salters Horners Advanced Physics book
